Read die following information carefully and answer the questions given below, A rectangular box is to be made from a sheet of 24 inch length and 9 inch width cutting out identical squares of side length x from ihe four corners and turning up the sides.
What is the value of x for which the volume is maximum?

Options

A. 1 inch
B. 1.5 inch
C. 2 inch
D. 2.5 inch

Answer

To find the value of \( x \) for which the volume of the rectangular box is maximized, we can follow these steps:

### Step 1: Define the dimensions of the box
We start with a rectangular sheet of dimensions 24 inches (length) and 9 inches (width). When we cut out squares of side length \( x \) from each corner and fold up the sides, the new dimensions of the box will be:
- Length: \( 24 - 2x \)
- Width: \( 9 - 2x \)
- Height: \( x \)

### Step 2: Write the volume function
The volume \( V \) of the box can be expressed as:
\[ V = \text{Length} \times \text{Width} \times \text{Height} = (24 - 2x)(9 - 2x)(x) \]

### Step 3: Expand the volume function
Now, we will expand the volume function:
\[ V = (24 - 2x)(9 - 2x)x \]
First, expand \( (24 - 2x)(9 - 2x) \):
\[ = 24 \cdot 9 - 24 \cdot 2x - 2x \cdot 9 + 4x^2 = 216 - 48x - 18x + 4x^2 = 216 - 66x + 4x^2 \]
Now, multiply by \( x \):
\[ V = x(216 - 66x + 4x^2) = 4x^3 - 66x^2 + 216x \]

### Step 4: Differentiate the volume function
To find the maximum volume, we need to differentiate \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = 12x^2 - 132x + 216 \]

### Step 5: Set the derivative equal to zero
To find critical points, set the derivative equal to zero:
\[ 12x^2 - 132x + 216 = 0 \]
Dividing the entire equation by 12 simplifies it:
\[ x^2 - 11x + 18 = 0 \]

### Step 6: Factor the quadratic equation
Now, we can factor the quadratic:
\[ (x - 9)(x - 2) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = 9 \]

### Step 7: Determine the feasible value of \( x \)
Since \( x \) represents the side length of the square cut from the corners, it must be less than half the width of the box. The maximum feasible value for \( x \) is:
\[ x < \frac{9}{2} = 4.5 \]
Thus, \( x = 9 \) is not a valid solution. Therefore, we take \( x = 2 \).

### Step 8: Conclusion
The value of \( x \) for which the volume of the box is maximized is:
\[ \boxed{2 \text{ inches}} \]